下面____程序段不能正确显示1!、2!、3!、4!的值。 A: For I=1 To 4 n=1 For j=1 to I n=n*j Next j Print n Next I B: For I=1 To 4 For j=1 to I n=1 n=n*j Next j Print n Next I C: n=1 For j=1 to 4 n=n*j Print n Next j D: n=1:j=1 Do While j<=4 n=n*j Print n j=j+1Loop

下列程序段不能正确显示1！、2！、3！、4！的结果的是 。 A: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox( n)Next i B: Dim i%, j%, n%For i = 1 To 4 n = 1 For j = 1 To i n = n * j Next j MsgBox( n)Next i C: Dim j%, n%n = 1For j = 1 To 4 n = n * j MsgBox( n)Next j D: Dim j%, n%n = 1j = 1Do While j<=4 n=n*j MsgBox(n) j=j+1Loop E: Dim j%, n%n = 1j = 1Do n=n*j MsgBox(n) j=j+1Loop While j<=4

下列程序段能且只能显示1！、2！、3！、4！的值的是（ ）。 A: For i = 1 To 4 n = 1 For j = 1 To i n = n * j Next j MsgBox(n)Next i B: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox(n)Next i C: n = 1For i = 1 To 4 For j = 1 To 4 n = n * j MsgBox(n) Next jNext i D: n = 1j = 1Do While j <= 4 n = n * j j = j + 1LoopMsgBox(n)

下列程序段能分别正确显示1!、2!、3!、4!的值的是______。 A: For i = 1 To 4 n = 1 For j = 1 To i n = n * j MsgBox(n) Next jNext i B: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox(n)Next i C: n = 1For i = 1 To 4 For j = 1 To 4 n = n * j MsgBox(n) Next jNext i D: n = 1j = 1Do While j &lt;= 4 n = n * j MsgBox(n) j = j + 1Loop

执行下列程序段后，输出结果是 n = 0 For i=1 To 3 For j=5 To 1 Step -1 n = n + 1 Next j, i Debug.Print n; j; i

‌二维数组a有m行n列，则在a[i]&#91;j&#93;之前的元素个数为（）​[/i] A: j*n+i B: i*n+j C: i*n+j-1 D: i*n+j+1

‎二维数组a有m行n列，则在a[i]&#91;j&#93;之前的元素个数为‏[/i] A: i*n+j B: j*n+i C: i*n+j-1 D: i*n+j+1

‌二维数组a有m行n列，则在a[i]&#91;j&#93;之前的元素个数为‏[/i] A: i*n+j B: j*n+i C: i*n+j-1 D: i*n+j+1

在C语言中，如要给一个较大的[img=54x22]17de83d2fb35c85.png[/img]的数组赋值，以下方式性能较好的是：（） A: for ( int i = 0; i <; N; i++ ) for ( int j = N - 1; j >;= 0; j-- ) mat[i][j] = val; B: for ( int i = 0; i <; N; i++ ) for ( int j = 0; j <; N; j++ ) mat[j][i] = val; C: for ( int i = N - 1; i >;= 0; i-- ) for ( int j = N - 1; j >;= 0; j-- ) mat[j][i] = val; D: for ( int i = N - 1; i >;= 0; i-- ) for ( int j = N - 1; j >;= 0; j-- ) mat[j][i] = val;

试分析下面代码段的时间复杂度： for（i=1；i<=n；++i） for（j=1；j<=n；++j） { ++x； s+=x； } A: O(1) B: O(n) C: O(n^2) D: O（n^3）